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Let $H(n)$ be the group of self-homotopy-equivalences of $S^n$ preserving the basepoint. I read that $H(n)$ may be identified with ''two components of $\Omega^nS^n$''.

What does this mean and how can I see it?

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up vote 3 down vote accepted

Maps $[S^n,S^n]$ are classified by degree. For each $z\in\mathbb{Z}$ there is a component of $\Omega^nS^n$. The only maps that are homotopy equivalences have degree $1$ and $-1$.

EDIT: In answer to the comment below from JasonDevito: For any finite CW complex $X$, $[S^1,X]$ is the same thing as $[S^0,\Omega X]$, since $\Omega$ and $\Sigma$ (reduced suspension) are adjoints. But, based maps from $S^0$ to $Y$ is the same as $Y$ for any finite CW complex. Thus

$$ [S^1,X]=[S^0,\Omega X]=\Omega X. $$

Iterate this and you get

$$ [S^n,S^n]=[\Sigma S^{n-1},S^n]=[S^{n-1},\Omega S^n]=\cdots =\Omega^n S^n. $$

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I feel as though I'm being dense, but I have an intuitive way of thinking about $\Omega S^n$. How does one think about $\Omega^n S^n$ in terms of $[S^n,S^n]?$ For example, how does a loop of loops in $S^2$ correspond to a map from $S^2$ to itself? Is each loop coming from, say, a lattitude of $S^2$ and the north and south poles are, up to homotopy, sent to the same point (making a trivial loop?) – Jason DeVito Jul 3 '12 at 14:38
@JasonDeVito: Does my edit answer you question? – Joe Johnson 126 Jul 3 '12 at 20:57
It does - I was being very dense. I've actually used these facts on answers I've posted on Mathoverflow - I'm just running really low on sleep. Sorry to be a bother! – Jason DeVito Jul 3 '12 at 21:17
If these are homotopy classes then isn't $\mathbb Z = \pi_n(S^n) = [S^n, S^n] =[S^0, \Omega^n S^n ] = \pi_0(\Omega^n S^n)$, with the two path components corresponding to $1$ and $-1$. – Justin Young Jul 4 '12 at 21:57

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